Not sure where to go from here.

Question about the task

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Actual question today, not a proofreading error; I’m a little stumped. It says I need to remove “r” words but I did that, so I thought. I have a feeling it’s a logic order issue but it should work the output in the compiler is ok? Any help is gladly accepted,

1. Create a list of words and populate it yourself.
2. The fix method should:
2.1. remove all words containing the letter "r" from the list
2.2. duplicate all words containing the letter "l".
2.3. if a word contains both "r" and "l", then leave it unchanged.
2.4. don't do anything to other words.

For example:
rose lyre love

Output:
lyre love love

Requirements:

The program must not read data from the keyboard.
The fix method should remove all words containing the letter "r" from the list. There is an exception: words containing both "r" and "l" should be left alone.
The fix method must duplicate words containing the letter "l" (add another instance of this word to the list). There is an exception: words containing both "l" and "r" should not be duplicated.
The fix method should not do anything with words that contain both "l" and "r".
The fix method should not do anything with words that don't contain either "l" or "r".

package com.codegym.task.task07.task0716; import java.util.ArrayList; /* R or L */ public class Solution { public static void main(String[] args) throws Exception { ArrayList<String> list = new ArrayList<String>(); list.add("rose"); // 0 list.add("love"); // 1 list.add("lyre"); // 2 list = fix(list); for (String s : list) { System.out.println(s); } } public static ArrayList<String> fix(ArrayList<String> list) { for (int i = 0; i < list.size(); i++) { String s = list.get(i); if (s.contains("l") && (s.contains("r"))) { continue; } if (s.contains("l") && !(s.contains("r"))) { list.add(i+1,s); } if (s.contains("r") && !(s.contains("l"))) { list.remove(s); } System.out.println(list.get(i)); } return list; } }

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Nouser

Level 36 , Germany

29 July 2020, 06:49

The problem is changing the size of a collection within a loop. This usually leads to problems or even exceptions. Just imagine that the collection size is determined before the loop starts and not changed anymore later on. So if you remove one entry, the loop still will continue to the old size and exceed the new borders. That's why you usually should use an iterator in these cases. Of course you can solve the task with an additonal counter variable or with modifying the loop by hand as often seen in the help section here. It's up to you what you like more. The iterator at the beginning may seem more complex but if you get used to it, it's very comfy. Unfortunately the task here is double complex with removing and adding elements at certain indexes, cause over a standard iterator adding is not possible. Here you can use the ListIterator. It is used exactly as a regular iterator but has additional features like adding elements. If you code the version with keeping track of the index on your own, you of course can do everything in one loop. You should just keep in mind to increase the index by one if there's no match, if you have to remove an entry, do not change the index and if you add an element, increase the index by two.